# A laplacian working on an equation containing a laplacian and a gradient

I have an equation as follows:

$$a \Delta \mathbf{u} + \mathbf{\nabla}(\mathbf{\nabla} \cdot \mathbf{u}) = 0$$

in which $$a$$ is a constant, $$\mathbf{u}$$ is a vector, $$\Delta$$ is the Laplacian operator, $$\mathbf{\nabla}$$ is the gradient operator and $$\mathbf{\nabla} \cdot$$ is the divergence operator.

If I take a Laplacian of the above equation, the answer is supposed to be:

$$\Delta \Delta \mathbf{u} = 0.$$

But how could I reach this?

(Actually, these two equations are eq.(7.4) and eq.(7.7) in the book of Theory of Elasticity - 1970 written by L. Landau)

$$\bf{Edit.1}$$ - A possible solution

Since we know: $$\mathbf{\nabla}(\mathbf{\nabla} \cdot \mathbf{u}) = \Delta \mathbf{u} + \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{u})$$, if we insert this equation to the first equation and then apply Laplacian operator on it, we will have a following equation:

$$a \Delta \Delta \mathbf{u} + \Delta \Delta \mathbf{u} + \mathbf{\nabla} \cdot ( \mathbf{\nabla} ( \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{u}) ) ) = 0$$

where I used the fact that $$\Delta \mathbf{v} = \mathbf{\nabla} \cdot (\mathbf{\nabla} \mathbf{v} )$$, in which $$\mathbf{v}$$ is another vector.

Now, if we assume that we can commute $$\mathbf{\nabla}$$ with $$\mathbf{\nabla} \times$$, the above equation would become:

$$a \Delta \Delta \mathbf{u} + \Delta \Delta \mathbf{u} + \mathbf{\nabla} \cdot ( \mathbf{\nabla} \times ( \mathbf{\nabla} (\mathbf{\nabla} \times \mathbf{u}) ) ) = 0$$

Since it is known that $$\mathbf{\nabla} \cdot ( \mathbf{\nabla} \times \mathbf{v}) = 0$$, finally we will have:

$$(a + 1) \Delta \Delta \mathbf{u} = 0$$

Then, either we say $$a = -1$$, or we say $$\Delta \Delta \mathbf{u} = 0$$.

It turns out that $$a = 1- 2 \sigma$$, in which $$\sigma$$ represents the Poisson's ratio. For a stable, isotropic and linear elastic material, $$\sigma$$ must be between -1.0 and 0.5. Thus, the possibility of $$a = -1$$ can be ruled out, and we have $$\Delta \Delta \mathbf{u} = 0$$ at the end.

Anyway, the possibility of the interchange of the gradient and curl would be critical for my approach.

$$\bf{Edit.2}$$ There is a problem in Edit.1

The curl of a gradient is 0, but it is not necessary that the gradient of a curl is also 0. Thus, the assumption that I made in Edit.1, which is that we can commute the curl and gradient, is wrong!

• Is there any statement made about the coordinate system? In cartesian coordinates, the Vector Laplacian boils down to $\boldsymbol \Delta \boldsymbol u = \Delta u_i$. Otherwise, I am not sure if you can prove this since you just bite always your own tail if you play around with the vector identities. Commented Apr 6, 2022 at 8:03
• @DanDoe Since this is about a general statement of the property of displacements in the theory of elasticity, I think a 3D Cartesian coordinates is OK. Commented Apr 6, 2022 at 11:56

## 1 Answer

The problem can proved by this way.

If we take a divergence of the first equation, we get

$$\Delta (\mathbf{\nabla} \cdot \mathbf{u}) = 0$$

Now, if we take the Laplacian of the first equation and take into account that a Laplacian and a gradient is commutable, we get

$$a \Delta \Delta \mathbf{u} + \mathbf{\nabla}[\Delta (\mathbf{\nabla} \cdot \mathbf{u})] = 0$$

then it is easy to see

$$\Delta \Delta \mathbf{u} = 0$$